Question

Honda Activa accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.

Answer 1

Answer:

0.71 m/s²

Explained:

Honda Activa was initially at rest, where its velocity was 0.

Later on, when it had covered 35.4 m, velocity becomed 7.10.

From this,

Initial velocity = 0

Final velocity = 7.10 m/s

Displacement = 35.4 m

Using equations of motion:

• v^2 = u^2 + 2aS

= > (7.10)² = 0² + 2a(35.4)

= > 50.41 = 70.8a

= > 50.41/70.8 = a

= > 0.71 m/s² ≈ a

Answer 2

Answer:

0.71 m/s²

Explanation:

Given; initial velocity is 0 m/s, final velocity is 7.10 m/s and distance covered is 35.4 m.

To find: Acceleration of the bike.

Using the third equation of motion,

v² - u² = 2as

Substitute the values,

→ (7.10)² - (0)² = 2a(35.4)

→ 50.41 = 70.8a

Divide by 70.8 on both sides,

→ 50.41/70.8 = 70.8a/70.8

On solving we get,

→ 0.71 m/s² = a

Hence, the acceleration of the bike is 0.71 m/s².